Phương trình vi phân cấp n

F y

y x
F y
−
p−
C−
−
F (x, y, y

,y

, ,y
(m)
)=0
y = y(x)
y = y(x)
I ⊂ R F
G R × R
m+1
y(x)=(y
1
(x), ,y
n
(x))
T
F
R
n
m m

F (x, y, y

)=0
F (x, y, z)
G ⊂ R
3
y

= f(x, y)
f D ⊂ R
2
e
y
+ y

2
cos x =1
y

2
− 2xy =lnx
∂
2
u
∂x
2
+
∂
2
u

∂y
2
=0
y : I → R
n
I =(a, b)
R m
I
F (x, y(x),y

(x),y

(x), ,y
(m)
)(x)=0 x ∈ I
y = y(x)
y = C
1
cos x + C
2
sin x
y

+ y =0
x =
x(t) y = y(t)
−
y

= y(α − βx),x


= x(γy − δ)
α, β, γ δ
y x
dy
dx
=
y(α −βx)
x(γy − δ)
(γy − δ)
y
dy =
(α − βx)
x
dx
γy − δ ln y = α ln x − βx+ C
C α = β =
γ =1,δ =2
1234
1
2
3
yy
z
z
X
−
y =
x
3

3
+ C
y

= x
2
y =
x
3
3
+1 y(0) = 1
y(x)
y(x
0
)=y
0
(x
0
,y
0
) ∈ D
y

= f (x, y),y(x
0
)=y
0
f D ⊂ R
2
y(x)

y(x)
y(x)=y
0
+

x
x
0
f(t, y(t))dt
C
1
I
− ¨o
I
− ¨o
y
0
(x)=y
0
y
k+1
(x)=y
0
+

x
x
0
f(t, y
k

(t))dt, k ∈ N
f
D = {(x, y)/|x − x
0
|≤a, |y −y
0
|≤b}
M := max
(x,y)∈D
|f(x, y)| h := min

a,
b
M

x ∈ I := [x
0
−
h, x
0
+ h]
|y
k
(x) −y
0
|≤b, k
y
k
D
x

0
− h ≤ x ≤ x
0
+ h
|y
k
− y
0
| =





x
x
0
f(t, y
k−1
(t))dt




≤

x
x
0
|f(t, y

k−1
(t))|dt ≤ M |x −x
0
|≤Mh ≤ b

y

= −y
2
y(0) = 1
y =
1
x +1
¨o y
0
=1 y
1
=1− x
y
2
=1−x + x
2
−
x
3
3
y
k
x
x

1234
Y (x)
2
Y (x)
0
Y (x)
4
Y (x)
1
Y (x)
3
− y

= −y
2
y(0) = 1
f(x, y) D ⊂ R
2
f
D y L
|f(x, y
1
) − f(x, y
2
)|≤L |y
1
− y
2
|, (x, y
1

), (x, y
2
) ∈ D
∂f
∂y
D
∂f
∂y




∂f
∂y




≤ M
f(x, y) y
f(x, y
1
) − f(x, y
2
)=(y
1
− y
2
)
∂f

∂y
[x, y
1
+ θ(y
2
− y
1
)]
f(x, y)
y
D = {(x, y)/ |x − x
0
|≤a, |y − y
0
|≤b}
I :=
[x
0
− h, x
0
+ h] h := min(a,
b
M
) M := max
(x,y)∈D
|f(x, y)|
I
|y
k+1
(x) −y

k
(x)|≤ML
k
|x −x
0
|
k+1
(k +1)!
, x ∈ I
k =0




x
x
0
f(t, y
k−1
(t))dt



≤ M |x − x
0
|
k − 1 x
0
≤ x ≤ x
0

+ h
|y
k+1
(x) − y
k
(x)| =





x
x
0
[f(t, y
k
(t)) − f(t, y
k−1
(t))] dt




≤

x
x
0
|f(t, y
k

(t)) − f(t, y
k−1
(t))|dt ≤ L

x
x
0
|y
k
(t) −y
k−1
(t)|dt
≤ L

x
x
0
|y
k
(t) − y
k−1
(t)|dt
≤ ML
k

x
x
0
|x −x
0

|
k
k!
dt = ML
k
|x −x
0
|
k+1
(k +1)!
x
0
− h ≤ x ≤ x
0
{y
k
(x)} I
|y
k+p
(x) − y
k
(x)|≤|y
k+p
(x) −y
k+p−1
(x)| + |y
k+p−1
(x) − y
k+p−2
(x)| + ···+ |y

k+1
(x) −y
k
(x)|
≤
M
L

(L |x − x
0
|)
k+p
(k + p)!
+ ···+
(L |x − x
0
|)
k+1
(k +1)!

≤
M
L

j≥k+1
(Lh)
j
j!

∞

j=0
(Lh)
j
j!
k {y
k
(x)}
I y(x) y(x)
y
k+1
(x)=y
0
+

x
x
0
f(t, y
k
(t))dt
{y
k
(x)} f D
{f(t, y
k
(t))} I f(t, y(t))
y(x)
z(x)
y(x) −z(x)=


x
x
0
[f(t, y(t)) − f(t, z(t))] dt
|y(x) −z(x)| =





x
x
0
[f(t, y(t)) − f(t, z(t))] dt




≤ 2M |x −x
0
|
k
|y(x) − z(x)|≤2ML
k
|x − x
0
|
k+1
(k +1)!
, x ∈ I

k −→ +∞ |y(x) − z(x)| =0 I
y(x) 
f(x, y) R
2
y

=2

|y|,y(0) = 0
y ≡ 0
y(x)=

(x −C)
2
x ≥ C
0 x ≤ C
y(x)=

0 x ≥ C
−(x − C)
2
x ≤ C
E d T : E → E
α ∈ (0, 1) x, y ∈ E
d(Tx,Ty) ≤ αd(x, y)
T
x
∗
∈ E
T (x

∗
)=x
∗
123-3 -2 -1
1
-1
y

=2

|y|,y(0) = 0
y = y(x) y = y(x)
(x
0
,y
0
)
(x
0
,y
0
) ∈ D
D ⊂ R
2
D y = y(x, C) C
(x
0
,y
0
) ∈ D C

C = ϕ(x
0
,y
0
)(∗)
ϕ
y = y(x, C) C (∗)
(x
0
,y
0
) D
ϕ(x, y)=C ϕ(x, y)
y

+ y =0 y(x)=Ce
−x
C
•
•
(x
0
,y
0
) ∈ D C
0
= ϕ(x
0
,y
0

) y = y(x, C
0
)
y(x) y

=3y + x y(0) = 1
y = −
x
3
−
1
9
+Ce
3x
C
1=y(0) = −
1
9
+ Ce
0
C =
10
9
y = −
x
3
−
1
9
+

10
9
e
3x
f(x, y) R
2
M(x, y)
k =
dy
dx
= f (x, y)
y = ϕ(x, C)(∗)
C
x
dy
dx
=
∂ϕ
∂x
(x, C)
(∗) (x, y) C = C(x, y)
C
y

=
∂ϕ
∂x
(x, C(x, y)) =: f(x, y)
–3
–2

–1
1
2
y(x)
–3 –2 –1 1 2 3
x
y

= −
y
x
y = Cx
2
x y

=2Cx C
y

=2
y
x
M(x)dx + N(y)dy =0
M(x),N(y)

M(x)dx +

N(y)dy = C
y
2
y


= x(1 + x
2
)
y
2
dy −x(1 + x
2
)dx =0
y
3
3
−
x
2
2
−
x
4
4
= C
M
1
(x)N
1
(y)dx + M
2
(x)N
2
(y)dy

M
2
(x)N
1
(y)
0
M
1
(x)
M
2
(x)
dx +
N
2
(y)
N
1
(y)
dy =0

M
1
(x)
M
2
(x)
dx +

N

2
(y)
N
1
(y)
dy = C
x(1 + y
2
)dx + y(1 + x
2
)dy =0
(1 + x
2
)(1 + y
2
)
xdx
1+x
2
+
ydy
1+y
2
=0

xdx
1+x
2
+


ydy
1+y
2
= C
1
2
ln(1 + x
2
)+
1
2
ln(1 + y
2
)=C :=
1
2
ln C
1
(1 + x
2
)(1 + y
2
)=C
1
C
1
f(x, y) m t>0
f(tx, ty)=t
m
f(x, y)

y

= f(x, y)
0 f(tx, ty)=f(x, y)
t>0
u :=
y
x
f(x, y)=f(±|x|, |x|
y
|x|
)=f(±1, ±u)=:g(u)
y = xu
dy
dx
= x
du
dx
+ u
x
du
dx
+ u = g(u)
du
g(u) − u
=
dx
x

du

g(u) − u
=ln



x
C



x = C exp

du
g(u) − u
C =0
u =
y
x
(x
2
+ y
2
)dx + xydy =0
dy
dx
= −
y
x
−
x

y
y = xu x
du
dx
+ u + u +
1
u
=0
dx
x
= −
udu
1+2u
2
ln



x
C



= −
1
4
ln(1 + 2u
2
)
u =

y
x
x
4
=
C
4
x
2
x
2
+2y
2
C =0
dy
dx
= f(
ax + by + c
a
1
x + b
1
y + c
1
)

x = ξ + x
0
y = η + y
0

x
0
y
0

ax
0
+ by
0
+ c =0
a
1
x
0
+ b
1
y
0
+ c
1
=0
dη
dξ
= f

aξ + bη
a
1
ξ + b
1

η

= f

a + b
η
ξ
a
1
+ b
1
η
ξ

= g

η
ξ

(2x − 4y +6)dx +(x + y − 3)dy =0

2x
0
− 4y
0
+6=0
x
0
+ y
0

− 3=0
x
0
=1,y
0
=2

x = ξ +1
y = η +2
(2ξ − 4η)dξ +(ξ + η)dη =0
η = uξ
(2 − 3u + u
2
)dξ + ξ(1 + u)du =0
u =1 u =2
2 − 3u + u
2
dξ
ξ
+
(1 + u)du
2 − 3u + u
2
=0
⇔
dξ
ξ
+

3

u − 2
−
2
u − 1

du =0
ln |ξ| +ln
|u − 2|
3
(u − 1)
2
=lnC
1
ξ
(u − 2)
3
(u − 1)
2
= C
x, y
(y −2x)
3
= C(y − x −1)
2
y = x +1 y =2x u =1 u =2
P (x, y)dx + Q(x, y)dy =0
U(x, y)
dU(x, y)=P (x, y)dx + Q(x, y)dy
U(x, y)=C
∂P

∂y
=
∂Q
∂x
U(x, y)
U(x, y)=

x
x
0
P (x, y)dx +

y
y
0
Q(x
0
,y)dy
U(x, y)=

x
x
0
P (x, y
0
)dx +

y
y
0

Q(x, y)dy
(x
0
,y
0
)
(x
3
+ xy
2
)dx +(x
2
y + y
3
)dy =0
P (x, y)=x
3
+ xy
2
Q(x, y)=x
2
y + y
3
∂P
∂y
=2xy =
∂Q
∂x
U(x, y)
U(x, y)=


x
0
(x
3
+ xy
2
)dx +

y
0
(0.y + y
3
)dy
U(x, y)=
x
4
4
+
x
2
y
2
2
+
y
4
4
(x
2

+ y
2
)
2
=4C
1
:= C
2
x
2
+ y
2
= C C ≥ 0
µ(x, y)
µ(x, y){P (x, y)dx + Q(x, y)dy} =0
µ(x, y)
µ µ
∂
∂y
(µP )=
∂
∂x
(µQ)
Q
∂µ
∂x
− P
∂µ
∂y
= µ


∂P
∂y
−
∂Q
∂x

(∗)
µ
µ x
µ>0 (∗) µ
d ln µ
dx
=
∂P
∂y
−
∂Q
∂x
Q
=: ϕ
y
µ(x)=exp


ϕ(x)dx

µ y
µ(y)=exp



ψ(y)dy

ψ(y):=
∂Q
∂x
−
∂P
∂y
P
x
(2xy+x
2
y+y
3
/3)dx+(x
2
+y
2
)dy =
0
P (x, y)=2xy + x
2
y + y
3
/3 Q(x, y)=x
2
+ y
2
∂P

∂y
−
∂Q
∂x
Q
=
2x + x
2
+ y
2
− 2x
x
2
+ y
2
=1
µ(x)=exp(

dx)=e
x
e
x
[(2xy + x
2
y + y
3
/3)dx +(x
2
+ y
2

)dy]=0
ye
x
(x
2
+ y
2
/3) = C
y

+ p(x)y = q(x)
p(x),q(x) (a, b)
q(x) ≡ 0
y

+ p(x)y =0
y y = u(x)v(x)
u

v + uv

+ p(x)uv = q(x)
v
v

+ p(x)v =0
dv
v
= −p(x)dx
ln |v| = −


p(x)dx +ln|C
1
|, C
1
=0
|v| = |C
1
|exp

−

p(x)dx

v =0
v(x)=Ce
− p(x)dx
v(x)=e
− p(x)dx
u

v = f(x)
u =

q(x)
v(x)
dx + C
u, v y
y = e
− p(x)dx



q(x)e
p(x)dx
dx + C

C
y

+3xy = x (0, 4)
p(x)=3x

p(x)dx =3x
2
/2
y = e
−3x
2
/2


xe
3x
2
/2
dx + C

= e
−3x
2

/2

1
3
e
3x
2
/2
+ C

=
1
3
+ Ce
−3x
2
/2
x =0 y =4 C =
11
3
y =
1
3
+
11
3
e
−3x
2
/2

y

+ p(x)y = y
α
g(x)
α
α =0 α =1
α =0 α =1
z = y
1−α
z

=(1− α)y
−α
y

y
−α
z z

z
z

+(1− α)p(x)z =(1−α)g(x)
y =0
y
α
xy

− 4y = x

2
√
y
α =1/2 y =0
y =0 xy
1/2
y
−1/2
y

−
4
x
y
1
2
= x
z = y
1
2
z

=
1
2
y
−1/2
y

z


−
2
x
z =
x
2
z = x
2

1
2
ln |x| + C

y = x
4

1
2
ln |x| + C

2
y =0
A(x, y)dx + B(x, y)dy + H(x, y)(xdy − ydx)=0
A, B m H n
n = m − 1
y = z.x
dy = xdz + zdx, xdy − ydx = x
2
d


y
x

= x
2
dz
x
m
A(1,
y
x
)dx + x
m
B(1,
y
x
)dy + x
n
H(1,
y
x
)x
2
d

y
x

=0

x
m
[A(1,z)+zB(1,z)] dx +

xB(1,z)+H(1,z)x
n+2−m

dz =0
xB(1,z)+H(1,z)x
n+2−m
=0
dx
dz
+
B(1,z)
A(1,z)+zB(1,z)
x = −
H(1,z)
A(1,z)+zB(1,z)
x
n+2−m
x = x(z) z
xdx + ydy + x
2
(xdy − ydx)=0
y = xz
xdx + xz(xdz + zdx)+x
4
dz =0
(1 + z

2
)dx +(xz + x
3
)dz =0
dx
dz
+
z
1+z
2
x = −
1
1+z
2
x
3
1
x
2
= C(1 + z
2
)+(1+z
2
) arctan z + z
C(x
2
+ y
2
)+(x
2

+ y
2
) arctan

y
x

+ xy −1=0
C
−
y

= p(x)y
2
+ q(x)y + r(x)
p(x),q(x) r(x) (a, b)
p(x) ≡ 0
r(x) ≡ 0
˜y
˜y

= p(x)˜y
2
+ q(x)˜y + r(x)
y =˜y + z z
˜y

+ z

= p(x)˜y

2
+2p(x)˜yz + p(x)z
2
+ q(x)˜y + q(x)z + r(x)
z

− [2p(x)˜y + q(x)]z = p(x)z
2

y

+2y(y −x)=1
y = x
y = x + z
z

+2z(z + x)=0
α =2 u = z
−1
u

− 2xu =2
u = e
x
2


2e
−x
2

dx + C

y = x +
e
−x
2
C +2

e
−x
2
dx
, y = x
−
(xy
2
+4x)dx +(y + x
2
y)dy =0
2x

1 − y
2
+ yy

=0
y

= e
x+y

y

=
x
2
y − y
y +1
y

=
y
x
− 1
y

=
2xy
x
2
− y
2
(y
2
− 3x
2
)dy +2xydx =0
xy

= y ln
y

x
(x −2y +9)dx =(3x − y +2)dy
y

=2

y +2
x + y −1

2
y
x
dx +(y
3
+lnx)dy =0
e
y
dx +(xe
y
− 2y)dy =0
2xydx +(x
2
− y
2
)dy =0
[(x +1)e
x
− e
y
] dx = xe

y
dy
(x + y
2
)dx − 2xydy =0
(y
2
− 6xy)dx +(3xy − 6x
2
)dy =0
y(1 + xy)dx −xdy =0
xy ln ydx +(x
2
+ y
2

y
2
+1)dy =0
y

− 4y = x − 2x
2
xy

+ y = e
x